An Introduction to O-Minimality

An introduction to o-minimality

Notes by Tom Foster and Marcello Mamino on lectures given by Alex Wilkie at the MODNET summer school, Berlin, September 2007

1 Introduction

Throughout these notes definable means definable with parameters, unless otherwise stated. Definition 1. The structure R := hR, <, . . .i is said to be o-minimal if the only definable subsets of R are finite unions of intervals and points, ie. those sets definable with just the ordering.

Example 1. The real field R = hR, <, +, ·, 0, 1i is o-minimal. This is a conse- quence of Tarski’s theorem that R admits quantifier elimination. In fact we may conclude something stronger. Let φ(x) be a formula in the language of R with V W parameters from R. Writing φ(x) as i j pij(x)ij0, where ij ∈ {<, =} and pij ∈ R[X], we see that we have an upper bound, namely Σij deg(pij), on the set of points needed to describe φ(R) which is independent of the parameters used in φ. This is a special case of the following general property of definable families in o-minimal structures over the reals:

Theorem 1. Let R∗ be any o-minimal expansion of hR,

1.1 o-minimality via model completeness Deﬁnition 2. We say that a ﬁrst-order theory T is model complete if, modulo T , every formula is equivalent to an existential formula. When we say that a structure is model complete we mean that its theory is model complete.

1 Theorem 2 (Robinson’s test for model completeness). Let T be a ﬁrst-order theory. Suppose that T has the following property:

For any M, N T such that M ⊆ N , and any quantiﬁer-free formula φ(¯v) with parameters in M, if N ∃vφ¯ (¯v) then M ∃vφ¯ (¯v). Then T is model-complete.

Theorem 3. Let R∗ be an expansion of hR,

2. the theory of R∗ is model-complete. Then our structure R∗ is o-minimal. Proof. Let X be a definable subset of R∗. By the model-completeness of R∗ X is defined by an existential formula ∃yφ¯ (¯y, x), i.e. φ(¯y, x) is quantifier-free. The set, Y say, defined by φ(¯y, x) must have finitely many connected components. X is the image of Y under the projection map π : hy,¯ xi → x so must have finitely many connected components also. The only connected subsets of R are intervals and points so X is of the desired form.

Henceforth, R˜ := hR, Fi where F is some collection of functions from Rn to R, for various n. Deﬁnition 3. We call a term in the language of R˜ simple if has the form τ(¯x) = (1) (s) p(x1, . . . , xn,F1(¯x ),...,Fs(¯x )), wherex ¯ = (x1, . . . , xn), p(x1, . . . , xn, y1, . . . , ys) ∈ (i) R[x1, . . . , xn, y1, . . . , ys], F1,...Fs ∈ F and thex ¯ are subsequences ofx ¯ of suitable length. For example, if F = {exp} then a simple term is of the from

x1 xn p(x1, . . . , xn, e , . . . , e ).

Theorem 4. For R˜ := hR, Fi as above, every existential formula in the language of R˜ is equivalent, modulo the theory of R˜, to one of the form ∃y¯(τ(¯y, x¯) = 0) where τ is a simple term. Proof. Exercise. Hint: f(g(¯x)) = 0 iﬀ ∃y(y = g(¯x) ∧ f(y) = 0) iﬀ ∃y(f(y)2 + (g(¯x) − y)2 = 0)

2 Khovanski’s Theorem

Many early attempts to prove o-minimality were based on Khovanski’s ﬁniteness theorem.

Notation. For U ⊆ Rn open and m ∈ N ∪ {∞}, Cm(U) denotes the set of all functions f : U → R that are m-times continuously diﬀerentiable on U.

2 1 Deﬁnition 4. Let U ⊆ R. A sequence f1, . . . , fr ∈ C (U) is called a Pfaﬃan chain if there exist polynomials pi(x, y1, . . . , yi) ∈ R[x, y1, . . . , yi], for i = 1 . . . r, such that for all x ∈ U

0 f1(x) = p1(x, f1(x)) 0 f2(x) = p2(x, f1(x), f2(x)) . . 0 fr(x) = pr(x, f1(x), . . . , fr(x)).

n More generally, if U ⊆ R , we require polynomials pij(x1, . . . , xn, y1, . . . , yi) for j = 1, . . . , n, i = 1, . . . , r such that

∂fi (x1, . . . , xn) = pij(x1, . . . , xn, f(x1, . . . , xn), . . . , fi(x1, . . . , xn)). ∂xj

1 We say that a function f ∈ C (U) is Pfaffian if f(¯x) = q(¯x, f1(¯x), . . . , fr(¯x)) for some polynomial q and some Pfaffian chain f1, . . . , fr on U. The following facts about Pfaffian functions are left as exercises. Exercise 1. 1. If f ∈ C1(U) is Pfaffian then f ∈ C∞(U). 2. The class of Pfaffian functions includes many well-known and commonly occurring functions, for example exp on R, log on (0, ∞), sin on (0, π) and all rational functions on their domains. 3. The class of Pfaffian functions is closed under many operations, for example, paying close attention to domains, composition, differentiation, integration and extracting roots via the implicit function theorem.

Theorem 5 (Khovanski’s finiteness theorem). Let U ∈ Rm be an open box and suppose that f ∈ C∞(U) is Pfaffian. Then there exists a natural number n, depending only upon the degrees of the pij’s and the q as in definition 4, such that Z(f) := {x¯ ∈ U : f(¯x) = 0} has at most N connected components.

Exercise 2. Prove the following special case of theorem 5: Let U ⊆ R be an interval, and let f ∈ C∞(U) \ 0 satisfy f 0(x) = p(x, f(x)) for some p(x, y) ∈ R[x, y]. Prove that f has only finitely many simple zeroes. Find a bound n in terms of deg(p). Now use Sard’s Theorem to prove that f has only finitely many zeroes (approximate all zeroes by simple zeroes of a slightly different function). If you can extend this result to q(x, f(x)), for q(x, y) ∈ R[x, y], then you are well on the way to seeing the general case. Exercise 3. Prove that sin : R → R is not Pfaffian but sin : (0, π) → R is. What about sin : (0, 2π) → R? n Now let f1, . . . , fr be a Pfaffian chain on some open box U ⊆ R . Suppose ˜ ¯ we would like to show that R := hR, f1, . . . , fri is o-minimal (set fi(¯x) = 0 if x¯ ∈ / U). By theorems 3 and 5 it is sufficient that R˜ is model complete. In fact, we only know the model completeness of R˜ in a few cases:

3 x 1. n = r = 1, U = R and f1 = exp : x 7→ e .

2. arbitrary n, r but on bounded U such that f1, . . . fr have extensions, sat- isfying the same diﬀerential equations, to some open box V with U¯ ⊆ V .

However the o-minimality of R˜ is known in general, but by diﬀerent methods.

3 Real analytic functions

∞ Let U ⊆ R be an open interval and let f ∈ C (U). For x0 ∈ U, we form the (formal) Taylor series of f around x0:

(i) f ∞ f (x0) i T (x) = Σ (x − x0) (1) x0 i=0 i!

Then f is called analytic at x0 if there exists r > 0 (with (x0 − r, x0 + r) ⊆ U) such that for all y ∈ (x − r, x + r) the series of real numbers T f (y) converges 0 0 x0 with sum f(y). We say that f is analytic on U if it is analytic at x0 for all x0 ∈ U. Exercise 4. Let f ∈ C∞(U) where U is an open interval in R. Prove that the following are equivalent: 1. f is analytic on U. 2. for each compact set K ⊆ U there exists a constant C > 0 such that for all i ≥ 0 and for all x ∈ K, |f (i)(x)| ≤ Ci+1i! The deﬁnition of an analytic function generalizes easily to many variables once we have introduced multi-index notation. The result in exercise 4 also generalizes. One just replaces x, x0 by tuples, and i by a multi-index:-

n Notation. An element α = hα1, . . . , αni ∈ N is called a multi-index. We set

|α| := α1 + ... + αn

α! := α1! . . . αn! |α| (α) ∂ f f := α1 αn ∂x1 . . . ∂xn α α1 αn x¯ := x1 . . . xn wherex ¯ = (x1, . . . , xn)

We write Cω(U) for the set of analytic functions on U, where U is an open subset of Rn for some n. Fact 1. 1. The class of analytic functions is closed under composition, differentiation, integration and extracting roots via the implicit function theorem. 2. Every Pfaﬃan function is analytic on its domain.

4 3. If f is analytic on a connected open neighbourhood U of the point x0 and (α) n f (x0) = 0 for all α ∈ N then f ≡ 0 on U.

Definition 5. For n ∈ N \{0} we define to Gn to be the set of restrictions to [−1, 1]n of analytic functions defined on open neighbourhoods of [−1, 1]n, n n ie. f :[−1, 1] → R ∈ Gn iff there exists an open box V with [−1, 1] ⊆ V ω and g ∈ C (V ) such that f = g [−1,1]n . We let G0 := R and now define ¯ S Ran := hR, n≥0 Gi (we set our functions to zero outside the unit box).

We have the following theorems about the structure Ran.

Theorem 6 (Gabrielov (1969)). Ran is model complete.

Theorem 7 (Denef, van den Dreis [1]). hRan,Di admits quantiﬁer elimination, where x |x| ≤ |y| 6= 0 D : 2 → :(x, y) 7→ y R R 0 otherwise.

4 Some local theory of analytic functions

Deﬁnition 6. Let

• I be the interval [−1, 1] in R;

• Lan be the language which has, for each n ≥ 0 and each f ∈ Gn (see deﬁnition 5), a function symbol of arity n.

D • the language Lan be {D(·, ·), <} ∪ Lan; D • and, ﬁnally, I be the Lan-structure deﬁned on the domain I by interpreting D any function symbol in Lan as the analytic function it names, D as ( x if |x| ≤ |y| and 0 < |y| D(x, y) = y 0 otherwise,

and < in the usual way (moreover we will use >, ≤ and ≥ as shorthands for the corresponding deﬁnitions in terms of < and =). We are interested in proving the following Theorem 8. The structure I has quantiﬁer elimination. This implies theorem 7.

Proof that theorem 8 implies theorem 7. Let L be the language of our hRan,Di. Observe that hRan,Di can be interpreted in I by mapping each x ∈ R to the 2 a unique pair (a, b) ∈ I such that b = x and either a or b (or both) is 1. Moreover D each quantifier free Lan-formula is, ipso facto, an L-formula. Hence, suppose we want to eliminate the quantifiers in some L-formula φ(x1, . . . , xn): we first

5 D consider its interpretation as an Lan-formula ψ(a1, b1, . . . , an, bn), then, by the- D 0 orem 8, we have an equivalent quantiﬁer free Lan-formula ψ (a1, b1, . . . , an, bn), and, ﬁnally, we notice that φ is equivalent in hRan,Di to an appropriate boolean 0 combination of the formulæ|xi| ≤ 1 and instances of ψ with ai and bi replaced respectively by either xi and 1 or D(1, xi) and 1.

D Notice that, even if the symbols D and < in the language Lan are definable in Lan, we are not able to achieve the QE without them. Definition and facts First of all we will recall some definitions and facts about the real analytic functions.

n Definition 7. The ring Ox of germs of analytic functions at x is defined as the quotient of the ring of functions from Rn to R which are analytic on a neighbourhood of 0 by the equivalence relation ∼x defined by

f ∼x g ↔ ∃ > 0 ∀y |x − y| < → f(y) = g(y) i.e. identifying functions which are pointwise equal in some neighbourhood of x.

n Of course, O0 is isomorphic to the ring R hx1, . . . , xni of the power series in n variables convergent in some neighbourhood of 0. This observation enables us to state the following

Fact 2. The ring R [[x1, . . . , xn]] of formal power series in n variables is faith- n fully flat over the ring O0 . In particular, a linear equation with coefficients in n n O0 has a solution in O0 if and only if it has one in R [[x1, . . . , xn]]. We will need the Weierstrass preparation theorem. Let us first introduce the following definition.

n+1 Deﬁnition 8. Let f ∈ O0 . We say that f is regular of degree d in the variable xn+1 if f has non-vanishing dth derivative in xn+1 at 0 and zero ith derivative in xn+1 at 0 for all i < d. We say that f is regular if it is regular of degree d for some d.

n+1 Theorem 9 (Weierstrass preparation theorem). Let f(x1, . . . , xn+1) ∈ O0 be a germ of analytic functions which is regular of degree d in xn+1. Then

f = u · p

n+1 where u is a unit of O0 , and p is a monic polynomial in the indeterminate n xn+1 of degree d with coeﬃcients in O0 . Overview of the proof In the proof we will often deal with terms and formulæ in two sets of variables. We will write them as φ(¯x, y¯), wherex ¯ represents an m-tuple of variables andy ¯ represents an n-tuple of variables. Moreover we will call simple a formula or term in which no instance of D is applied to a term involving any of the

6 variables iny ¯. The symbols i and j will be used to denote multiindices in Nn, the norm |i| of a multiindex is the maximum of its components. We are now ready to prove the QE theorem. By a well known observation, D all we have to prove is that for any quantiﬁer free Lan-formula φ(¯x, y) exists a D quantiﬁer free Lan-formula ψ(¯x) such that

I ∃yφ(¯x, y) ↔ ψ(¯x) (wherex ¯ represents any number of free variables). D 0 Our strategy is to find a new quantifier free Lan-formula φ which is simple and satisfies 0 I ∃yφ(¯x, y) ↔ ∃yφ¯ (¯x, y¯) (thus eliminating the Ds at the expense of an increase in the number of existential quantifiers). Now we have a string of n existential quantifiers to eliminate, and suppose by induction that we can eliminate n−1 existential quantifiers (in front of a simple formula). Our goal is to reduce the formula ∃yφ¯ 0(¯x, y¯) to a boolean combination of simple formulæ each of which has at most n − 1 existential quantifiers, thus concluding by induction. 0 More specifically, assuming φ to be just an Lan-formula (a detail which has no influence on the argument overall), we will prove that for any (¯p, q¯) ∈ m n D I × I there exists an (¯p,q¯) > 0 and a simple quantifier free Lan-formula 00 φ(¯p,q¯)(¯x, z1, . . . , zn−1) such that

0 00 I ∃yφ¯ (¯p + (¯p,q¯)x,¯ q¯ + (¯p,q¯)y¯) ↔ ∃zφ¯ (¯p,q¯)(¯x, z¯) This means that for any point (¯p, q¯) ∈ Im × In we can get a simple formula which is equivalent to our one restricted to a neighbourhood of that point and has one quantifier less. Finally, we will conclude that, by compactness of Im × In, ∃yφ¯ 0(¯x, y¯) is equivalent to a boolean combination of just a finite number of such formulæ. 0 Of course, our assumption that φ is an Lan-formula can be easily fulfilled by replacing each instance of D with a dummy variable and then substituting the original terms back in φ00. Details of the proof The first step is easily concluded by replacing any (annoying) occurrence of D with its definition

2 2 2 2 I z = D(x, y) ↔ (x ≤ y → x = zy) ∧ ((x > y ∨ y = 0) → z = 0)

(in which all functions, i.e. square and product, are in Lan) existentially quan- tifying the dummy variable z. For the second step, we will work locally around (¯p, q¯). To simplify notation, we suppose, without loss of generality, that (¯p, q¯) = 0; moreover we make the assumption that each atomic subformula of φ0 is of the form t(¯x, y¯) > 0 with t a term in the language Lan. Our aim, now, is to ﬁnd a suitable value for = (¯p,q¯) as described above, thus ﬁlling in the missing part of the argument.

7 0 Of course, if an existentially quantified variable, say yn, occurs in φ just 0 polynomially (i.e. if each term of φ is a polynomial in the indeterminate yn whose coefficients are analytic functions in the remaining variables) then we can eliminate it by Tarski’s theorem (possibly multiplying each polynomial by a small enough constant in order to constrain the coefficients in I). Now, the Weierstrass preparation theorem provides us with a tool for making a variable occur just polynomially, at least locally. So, were each term in φ0 regular in yn , we could rewrite it locally around (¯p, q¯), which is around 0 by our assumption, so that yn occurs just polynomially. More specifically, if we can 0 m+n−1 write each term in φ as t = u·f locally at 0, with suitable f ∈ O0 [yn] and m+n unit u ∈ O0 , supposing without loss of generality u > 0 in a neighbourhood of 0, then we have for a small enough > 0

I t(x,¯ y¯) > 0 ↔ f(x,¯ y¯) > 0 which is suﬃcient to conclude by taking a value of small enough to work for every term in φ0, and invoking Tarski’s theorem as per the previous observation. Hence, the rest of this section will be devoted to reﬁning this argument in order to deal with the general case (when some of the terms may not be regular in yn+1 at (¯p, q¯), which we assumed to be 0).

m+n Lemma 1. For any f(¯x, y¯) ∈ O0 there is a positive integer d such that f(¯x, y¯) can be written as

X i f(¯x, y¯) = ai(¯x)¯y ui(¯x, y¯) |i|

m m+n with ai ∈ O0 and ui units of O0 . Proof. By induction on n. P k Case n = 1. Write the power series f(¯x, y) = k ak(¯x)y with ak(¯x) ∈ R hx1, . . . , xmi. Since R [[x1, . . . , xm]] is noetherian, there is an integer d such that the ideal generated by all the ak(¯x) is already generated by the ak(¯x) with k < d. Hence we can ﬁnd the units uk in R [[x1, . . . , xm, y]], and, by fact 2, even in R hx1, . . . , xm, yi. Case n > 1 By induction we can write

X i f(¯x, y¯) = ai(x, y1)(y2, . . . , yn) ui(¯x, y¯) |i| 0. By the P i lemma we can rewrite t as |i|

8 Let µj(¯x) be the formula   ^ aj(¯x) 6= 0 ∧  |aj(¯x)| ≥ |ai(¯x)| |i| 0 ↔ (µj(¯x) ∧ t(¯x, y¯) > 0) (*) |j|

(take j such that |aj(¯x)| is maximal, etc). We now focus on a single j, analyzing further the formula µj(¯x) ∧ t(¯x, y¯) > 0. Introducing the new variables vi, we deﬁne

j X i t˜(¯x, v,¯ y¯) =y ¯ uj(¯x, y¯) + viy¯ ui(¯x, y¯) |i|

0 t(¯x, y¯) = aj(¯x)t˜(¯x, v¯ (¯x), y¯)

0 where vi(¯x) = D(ai(¯x), aj(¯x)). We claim that after a suitable transformation of the variablesy ¯, the Weier- strass preparation theorem can be applied to t˜c¯ deﬁned by:

t˜c(¯x, v,¯ y¯) = t˜(¯x, c¯ +v, ¯ y¯)

n for anyc ¯ ∈ I({1,...,d}\{j}) . The transformation λ¯(¯y) is deﬁned by

( dn−k yk + yn if k 6= n λk(¯y) = yn if k = n and it has inverse ( dn−k yk − yn if k 6= n ωk(¯y) = yn if k = n ¯ our claim is that t˜c¯(¯x, v,¯ λ(¯y)) is regular in yn. Indeed, consider the lexicographically smallest multiindex ι such thaty ¯ι has non null coeﬃcient in the power series of t˜c¯(0, 0, y¯): since ui is a unit for every i, |ι| < d (in fact, ι either is j or lexicographically smaller than j with cι 6= 0). Now, the series ˜ ¯ P n−k of tc¯(0, 0, λ(0,..., 0, yn)) has order precisely n=1,...,n ιkd , since for any i lexicographically larger than ι

X n−k X n−k ιkd < ikd n=1,...,n n=1,...,n

9 so it is regular. By the Weierstrass preparation theorem ¯ t˜c¯(¯x, v,¯ λ(¯y)) = uc¯(¯x, v,¯ y¯) fc¯(¯x, v,¯ y¯)

whatever n with fc¯ ∈ O0 [yn] and unit uc¯ > 0 (and whatever = m + n + d − 2). Hence t˜c¯(¯x, v,¯ y¯) = uc¯(¯x, v,¯ ω¯(¯y)) fc¯(¯x, v,¯ ω¯(¯y)) and, by taking a function in each germ, the equality holds as well in some neighborhood Ic¯ of 0. At this point, for some c¯ > 0 we have the equivalence

¯ 0 t(¯x, λ(¯y)) > 0 ↔ aj(¯x)fc¯(¯x, v¯ (¯x) − c,¯ y¯) > 0

0 whenever µj(¯x) is true and the absolute values of vi(¯x) − ci, xk and yl (for all i, k and l) are all smaller than c¯. Observing that yn occurs just polynomially in 0 aj(¯x)fc¯(¯x, v¯ (¯x)−c,¯ y¯), possibly using the already mentioned trick of multiplying each term by a small enough constant, we may consider the right hand side of D the equivalence a simple L -formula in which yn occurs just polynomially. an n Now the dependency onc ¯ can be easily eliminated by compactness of I{1,...,d} \{j} 0 D and observing that |vi(¯x) − ci| < c¯ is (equivalent to) an Lan-formula. More precisely, for some > 0 and some ﬁnite set C of multiindices, whenever µj(¯x) is true and the absolute values of xk and yl (for all k and l) are all smaller than , t(¯x, λ¯(¯y)) > 0 is equivalent to

^ 0 ¯0 |vi(¯x) − ci| < c¯ → fc¯(¯x, v (¯x) − c,¯ y¯) > 0 c¯∈C

D which, again, is a simple Lan-formula in which yn occurs just polynomially. In the end, substituting the former in (*) for each j, for each atomic sub ¯ 0 ¯ D formula t(¯x, λ(¯y)) > 0 of φ (¯x, λ(¯y)) we have a simple Lan-formula in which yn occurs just polynomially equivalent to it in a neighbourhood of 0. Substi- tuting them and invoking Tarski’s theorem we accomplish our goal of locally eliminating one existential quantiﬁer.

5 The full exponential function

We aim to show that the real ﬁeld expanded by the full exponential function, which we will denote by Rexp, is model-complete. As remarked in section 2 we can then conclude that Rexp is o-minimal.

Definition 9. An expansion R˜ of the real field is said to be polynomially bounded if for all definable functions f : R → R there exists a natural number N such that |f(x)| ≤ xN for all sufficiently large x.

10 Remark 2. One might think that the full exponential function is somehow definable in Ran. If this were the case then the o-minimality of Rexp would follow directly from the o-minimality of Ran. However this is not the case. Indeed, in [4] van den Dries proves that the structure Ran is polynomially bounded, and hence cannot define the full exponential function. There are, however, some global analytic functions definable in Ran.

Example 2. sin (−π/2,π/2) and cos (−π/2,π/2) are both definable in Ran. Con- −1 sequently tan (−π/2,π/2) is definable in Ran. We may define tan on R by −1 saying that it is the inverse map to tan. So hR, tan i is o-minimal.

As remarked above Ran is polynomially bounded and so hR, exp [−1,1]i is polynomially bounded. We now develop some theory of polynomially bounded o-minimal expansions of the real ﬁeld. From now on, R˜ will denote such a structure.

Theorem 10 (Miller [2]). Let f : R → R be definable in R˜. Then there exists f(x) a unique α ∈ R such that xα converges to a non-zero finite limit as x → ∞. Exercise 5. With f and α as above, prove that α and x 7→ xα : (0, ∞) → R are definable over the same set of parameters as f. Furthermore, prove that the set of all such α forms a subfield of R; this is known as the field of exponents of R˜. From now we will assume that R˜ has field of exponents Q. This is the case ˜ ˜ ˜ for Ran [4] and hence for hR, exp [−1,1]i. Let T be the theory of R and let L be ˜ it’s language. Let M = hM, <, . . .i T . We define

µ(M) := {x ∈ M : ∀q ∈ Q>0 |x| < q} F in(M) := {x ∈ M : ∃q ∈ Q>0 s.t. |x| < q}. F in(M) is a subring of M and µ(M) is the unique maximal ideal of F in(M). In [5] van den Dries shows that the ﬁeld K := F in(M)/µ(M) can be expanded to an L˜ structure K˜ such that, up to isomorphism, K˜ can be elementarily embedded in M. In fact the embedding can be chosen so that each µ(M) equivalence class of F in(M) contains exactly one element of K˜ . Now F in(M)\µ(M) is a subgroup of the multiplicative group of M, M\{0}. We let Γ be the quotient

(M\{0}) /(F in(M)\µ(M)), which we will write additively, and we let v : M\{0} → Γ be the quotient map. Observe that since M is real closed Γ is divisible and hence a Q-vector space. Recall from [3] that the definable closure operator is a pregeometry in o- minimal structures so we have a notion of dimension. Furthermore, since o- minimal expansions of groups have definable Skolem functions, the dimension of our structure M is given by: inf{n : ∃n elements of M which generate M under 0-definable Skolem functions}.

11 Theorem 11 (The valuation inequaliy [6]). With M, K˜ and Γ as above, if dim(M) is finite then ˜ dim(M) ≥ dim(K) + dimQ(Γ). ˜ We also have a relative version of the valuation inequality: let K1,K2 T such ˜ ˜ that K1 4 K2. Let Γ1, Γ2 denote the value groups of K1, K2 and K1, K2 be their residue fields. If dimK1 (K2) is finite then

dim (K ) ≥ dim (Γ /Γ ) + dim (K˜ ). K1 2 Q 1 2 K˜ 1 2

For our purposes we will only use the fact that dimK1 (K2) ≥ dimQ(Γ2/Γ1).

Unravelling the statement, this means: if dimK1 (K2) = p, and a1, . . . ap+1 ∈ K2 then there exists n1, . . . , np+1 ∈ Z, not all zero, and c ∈ K1 such that n1 np+1 ca1 . . . ap+1 ∈ F in(K2) \ µ(K2).

5.1 Proof of model completeness of Rexp

Let Texp denote the theory of Rexp. Let M1, M2 Texp and suppose that M1 ⊆ M2. We claim that M1 4∃1 M2 and hence that Texp is model complete.

Sketch proof of claim. Let F (x1, . . . , xn) be a simple term with parameters in M1 and assume that there exists a1, . . . , an ∈ M2 such that F (a1, . . . , an) = 0. By Robinson’s test and theorem 4 it is suﬃcient to prove that there exists b1, . . . , bn ∈ M1 such that F (b1, . . . , bn) = 0.

x1 xn Step 1 F (x1, . . . , xn) is of the form P (x1, . . . , xn, e , . . . , e ) where P (¯x, y¯) ∈ 0 n 0 M1[¯x, y¯]. We construct P1,...,Pn ∈ M1[¯x, y¯] anda ¯ ∈ M2 such thata ¯ is a non-singular zero of the system of equations

x¯ Pi(¯x, e ) (i=1,. . . ,n) (2)

and a zero of P (¯x, ex¯). (3)

Step 2 We prove that ifa ¯0 is a non-singular of the system (2) which is bounded 0 between elements of M1 (ie. there exists b ∈ M1 such that |ai| < b for 0 i = 1, . . . n) then in facta ¯ lies in M1. Step 3 It now remains to prove that any non-singular zero of the system (2) in M2 is bounded between elements of M1. This is where we will use the valuation inequaltiy and the model completeness of hR, exp [0,1]i. Let hb1, . . . , bni be a non-singular solution of (2). We consider the structures ∗ ∗ M1 := hM1, exp [ −1, 1]i and M2 := hM2, exp [ −1, 1]i, where the bar indicates the ordered ﬁeld structure only. Let K2 be the elementary ∗ b1 bn ∗ substructure of M2 generated by b1, . . . , bn, e , . . . , e and M1. Then

12 dim ∗ K ≤ n, therefore by the valuation inequality, for each r = 1, . . . , n M1 2 ∗ there exists c ∈ M1 and a0, . . . , an ∈ Z not all zero such that

n a0 Y aibi cbr e ∈ F in(K2) \ µ(K2). i=1

0 0 By some combinatorial arguments we can now ﬁnd a1, . . . , an ∈ Z not ∗ Pn 0 all zero and α ∈ M1 such that −α ≤ i=1 aibi ≤ α. Now by a linear transformation of variables we may suppose that −α ≤ bn ≤ α. Now use an inductive argument.

References

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